What is (\partial)/(\partial y)(1/(t(y^2-9))) ?

The answer to (\partial)/(\partial y)(1/(t(y^2-9))) is -(2y)/(t(y^2-9)^2)

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(\partial)/(\partial y)(1/(t(y^2-9)))

\frac{\partial}{\partial y} (\frac{1}{t ( y ^{2} - 9 )})

- \frac{2 y}{t ( y ^{2} - 9 ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{1}{t ( y ^{2} - 9 )})

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{t} \frac{\partial}{\partial y} (\frac{1}{y ^{2} - 9})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{1}{t} \frac{\partial}{\partial y} ((y^{2} - 9)^{- 1})

Apply the chain rule:

- \frac{1}{( y ^{2} - 9 ) ^{2}} \frac{\partial}{\partial y} (y^{2} - 9)

= - \frac{1}{( y ^{2} - 9 ) ^{2}} \frac{\partial}{\partial y} (y^{2} - 9)

\frac{\partial}{\partial y} (y^{2} - 9) = 2 y

= \frac{1}{t} (- \frac{1}{( y ^{2} - 9 ) ^{2}} \cdot 2 y)

Simplify

\frac{1}{t} (- \frac{1}{( y ^{2} - 9 ) ^{2}} \cdot 2 y) : - \frac{2 y}{t ( y ^{2} - 9 ) ^{2}}

= - \frac{2 y}{t ( y ^{2} - 9 ) ^{2}}

f (x) = \frac{x}{9 - x ^{2}}, g (y) = 0, y = 2

\int e^{- i x} d x

x \to + 1 lim (∣ 2 x ∣)

\frac{d}{d x} (\frac{x ^{2} - 4}{6 x ^{3} - 2})

\frac{1}{2 x} + \frac{1}{4 x ^{\frac{3}{4}}}

What is (\partial)/(\partial y)(1/(t(y^2-9))) ?
The answer to (\partial)/(\partial y)(1/(t(y^2-9))) is -(2y)/(t(y^2-9)^2)